Introduction to Quantum Mechanics · Problem 6.18

Symmetries and Conservation Laws — David J. Griffiths

Physics — Quantum Mechanics · free_particle_degeneracy

Problem

Consider the free particle in one dimension: H^=p^2/2m\hat{H} = \hat{p}^2/2m. This Hamiltonian has both translational symmetry and inversion symmetry. (a) Show that translations and inversion don't commute. (b) Because of the translational symmetry we know that the eigenstates of H^\hat{H} can be chosen to be simultaneous eigenstates of momentum, namely fp(x)f_p(x) (Equation 3.32). Show that the parity operator turns fp(x)f_p(x) into f−p(x)f_{-p}(x); these two states must therefore have the same energy. (c) Alternatively, because of the inversion symmetry we know that the eigenstates of H^\hat{H} can be chosen to be simultaneous eigenstates of parity, namely 1π ℏcos⁡ ⁣(p xℏ)and1π ℏsin⁡ ⁣(p xℏ).\frac{1}{\sqrt{\pi\,\hbar}}\cos\!\left(\frac{p\,x}{\hbar}\right) \quad \text{and} \quad \frac{1}{\sqrt{\pi\,\hbar}}\sin\!\left(\frac{p\,x}{\hbar}\right). Show that the translation operator mixes these two states together; they therefore must be degenerate. *Note*: Both parity and translational invariance are required to explain the degeneracy in the free-particle spectrum. Without parity, there is no reason for fp(x)f_p(x) and f−p(x)f_{-p}(x) to have the same energy (I mean no reason based on symmetries discussed thus far ... obviously you can plug them in to the time-independent Schrödinger equation and show it's true).

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